Difference between revisions of "2024 AMC 12B Problems/Problem 13"
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==Solution 1 (Easy and Fast)== | ==Solution 1 (Easy and Fast)== | ||
− | Adding up the first and second | + | Adding up the first and second equation, we get: |
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | h + k &= 2x^2 + 2y^2 - 16x - 4y \ | ||
+ | &= 2(x^2 - 8x) + 2(y^2 - 2y) \ | ||
+ | &= 2(x^2 - 8x) + 2(y^2 - 2y) \ | ||
+ | &= 2(x^2 - 8x + 16) - (2)(16) + 2(y^2 - 2y + 1) - (2)(1) \ | ||
+ | &= 2(x - 4)^2 + 2(y - 1)^2 - 34 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | All squared values must be greater than or equal to <math>0</math>. As we are aiming for the minimum value, we set the two squared terms to be <math>0</math>. | ||
− | + | This leads to <math>\min(h + k) = 0 + 0 - 34 = \boxed{\textbf{(C)} -34}</math> | |
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− | This leads to (h+k) | ||
~mitsuihisashi14 | ~mitsuihisashi14 |
Revision as of 18:33, 14 November 2024
Contents
[hide]Problem 13
There are real numbers and that satisfy the system of equationsWhat is the minimum possible value of ?
Solution 1 (Easy and Fast)
Adding up the first and second equation, we get: All squared values must be greater than or equal to . As we are aiming for the minimum value, we set the two squared terms to be .
This leads to
~mitsuihisashi14
Solution 2 (Coordinate Geometry and HM-GM)
distance between 2 circle centers is min( h + k ) = .
Solution 3
~Kathan
See also
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.